For Spin 1/2 particles, the spin operator along an arbitrary axis defined by the normalized vector $\vec{n}$ is given by a weighted sum over the pauli matrices:

$$S(\vec{n})=n_x \sigma_x + n_y \sigma_y + n_z \sigma_z.\tag{I}$$

If a particle has been prepared in an up-eigenstate $\chi_n$ with respect to $S(\vec{n})$ and then we measure the spin again along a different axis $\vec{m}$, the probability that we will measure spin up again is given by the simple formula

$$|\langle \chi_m|\chi_n\rangle|^2=\cos\left(\frac{\theta}{2}\right)^2.\tag{II}$$

where $\theta$ is the angle between the two axes.

This formula can be derived by considering the general up-state eigenvector of $S(\vec{n})$, then calculating it`s inner product with another general eigenvector of $S(\vec{m})$ and then taking the absolute square, which is quite a mess of a calculation.

My question is: Is there an easy/elegant way to derive Eq. II that rests on as few assumptions as possible?


1 Answer 1


Since $|\langle \chi_m|\chi_n\rangle|^2$ is a scalar, we can without loss of generality take $\vec n$ to be along $+\hat z$ and $\vec m$ to be $R_z(\varphi)R_y(\theta)\vec n$. Then $\vert \vec n\rangle=\vert +\rangle$ and $\vert \vec m\rangle = R_z(\varphi)R_y(\theta) \vert +\rangle$. From this it immediately follows that $$ \langle + \vert \vec m\rangle = \langle + \vert R_z(\varphi)R_y(\theta) \vert +\rangle = e^{-i\varphi/2}\cos\left(\frac{\theta}{2}\right) $$ and your result follows immediately.

For a general orientation, you can replace $\vert +\rangle$ by $R(\omega)\vert +\rangle$ and $\vert \vec m\rangle $ by $R(\omega)\vert m\rangle$ with $R(\omega)$ an arbitrary rotation as this does not affect the relative angle between the vectors. Then $$ \langle + \vert R^{-1}(\omega)R(\omega)\vert \vec m\rangle = \langle +\vert \vec m\rangle\, , $$ and the general result follows.

Note that, for spin-1/2, $R_z(\varphi)=e^{-i\varphi\sigma_z/2}$ so $\langle +\vert R_z(\varphi)= e^{-i\varphi/2}\langle +\vert$ and that likewise $$ R_y(\theta)\vert +\rangle=e^{-i\theta \sigma_y/2}\vert +\rangle =\cos\left(\frac{\theta}{2}\right) \vert +\rangle + \sin\left(\frac{\theta}{2}\right) \vert -\rangle\, , $$ using elementary properties of $\sigma_y$.

For vectors we have $R_z(\varphi)=e^{-i\varphi L_z}$ and $R_y(\theta)=e^{-i\theta L_y}$ with $L_z$ and $L_y$ the $3\times 3$ matrices representing the appropriate generators.

  • $\begingroup$ So i guess the only thing we needed to assume here is how the rotation operator that transforms a 3D vector acts upon the corresponding ket vector?Is there a simple way to see that it has to be a multiplication by exp(-i phi/2)*C0s(th/2) ? Also, is it true that in the way you defined it, |+> = [1,0] (the eigenvector of sigma z)? $\endgroup$
    – curio
    Aug 18, 2017 at 19:39
  • $\begingroup$ @curiosity I added some additional details. $\endgroup$ Aug 18, 2017 at 19:43
  • $\begingroup$ @curiosity ... and yes $\vert +\rangle$ is [1,0] $\endgroup$ Aug 18, 2017 at 20:46

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